How is it possible for a King and Knight to checkmate a King and Pawn? I've composed a working position, but I am not sure if this can occur through a series of moves.


In the general case K+N vs K+P is of course a draw - or a win for the pawn if it can promote unhindered.

There is however a famous construction were the knight can force a mate against a king stuck in front of its own well-advanced rook pawn:

[fen "8/3N4/8/8/8/p7/k7/2K5 w - - 0 1"]

1.Nc5 Ka1 2.Kc2 Ka2 3.Nd3 Ka1 4.Nc1 a2 5.Nb3#

There are many variations on that theme, the first such position being published by in the Bonus Socius before 1300.

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  • why can't 2. Ka2 be 2. a2? Looks to me like that has a high chance of ending in stalemate. EDIT: ah, it's Nb3 checkmate – Cruncher Aug 7 at 13:18

Just for the record, the longest win in this endgame is 7 moves:

[FEN "8/8/8/8/p7/8/N7/k1K5 w - - 0 1"]

1. Nb4 a3 2. Nc2+ Ka2 3. Nd4 Ka1 4. Kc2 Ka2 5. Ne2 Ka1 6. Nc1 a2 7. Nb3#

The idea in this position (and other similar positions) is to stalemate the king in the corner; that forces a pawn move and when the pawn reaches a2, you'll need to have the knight positioned in such a way that it can immediately deliver mate.

Generally speaking, this only works (up to symmetry) if the pawn is on a3, the Black king on a1 or a2 and the White on c1 or c2 (otherwise, the king can escape) and the knight is both close enough (otherwise, Black can stalemate themselves) and has the right 'parity' - if Black to move loses, then White to move will only be able to draw. Likewise, if Black to move draws, then White to move wins. You can check a position in an online endgame tablebase like Sygyzy.

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It is not possible for king+knight to force checkmate but a helpmate is possible.

Basically the pawn promotes to a bishop, the king moves to the opposite coloured square to the bishop with the bishop next to it and the opposing knight delivers checkmate with the king blocking off the two squares in front of the other king.

Like this:

[fen "KB6/8/kn6/8/8/8/8/8 w - - 0 1"]
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  • 8
    This isn't the case - see Evargalo's answer for one example. – Inertial Ignorance Aug 6 at 20:35

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